1. ## Derivatives using ln

Got a question here....

Find the derivative of :

P(t) = 500 / (1 + e^-t)

I started by re-writing it in the form :

P(t) = 500 (1 + e^-t)^-1

Using the chain rule I came up with :

P'(t) = 500(-1)(1 + e^-t)^-2 (1 + e^-t)'
= -500(1 + e^-t)^-2(e^-t)(ln e)
= -500(1 + e^-t)^-2(e^-t)(1)

The question goes on to ask for the derivative at P'(3)

P'(3) = -500(1 + e^-3)^-2(e^-3)
= -500(.9074)(.04979)
= -22.6

I know the answer for this question should not be negative as it deals with population regrowth. Where did I go wrong? Thanks.

2. Originally Posted by Jools
Got a question here....

Find the derivative of :

P(t) = 500 / (1 + e^-t)

I started by re-writing it in the form :

P(t) = 500 (1 + e^-t)^-1

Using the chain rule I came up with :

P'(t) = 500(-1)(1 + e^-t)^-2 (1 + e^-t)'
= -500(1 + e^-t)^-2(e^-t)(ln e)

From where did you bring that $\displaystyle \ln e=1$ ? It can only confuse you.

= -500(1 + e^-t)^-2(e^-t)(1)

The question goes on to ask for the derivative at P'(3)

P'(3) = -500(1 + e^-3)^-2(e^-3)

Here's the mistake: you actually have $\displaystyle P'(3)=-500(1+e^{-3})(-2e^{-3})=1,000(1+e^{-3})e^{-3}$ , which is positive indeed.

Tonio

In fact now I realized that that -2 is a power (!!), but still: you wrote $\displaystyle (1+e^{-t})'=e^{-t}$ , which is lacking a minus sign there!

Tonio

= -500(.9074)(.04979)
= -22.6

I know the answer for this question should not be negative as it deals with population regrowth. Where did I go wrong? Thanks.
.

3. Originally Posted by Jools
Got a question here....

Find the derivative of :

P(t) = 500 / (1 + e^-t)

I started by re-writing it in the form :

P(t) = 500 (1 + e^-t)^-1

Using the chain rule I came up with :

P'(t) = 500(-1)(1 + e^-t)^-2 (1 + e^-t)'
= -500(1 + e^-t)^-2(e^-t)(ln e)
= -500(1 + e^-t)^-2(e^-t)(1)

The question goes on to ask for the derivative at P'(3)

P'(3) = -500(1 + e^-3)^-2(e^-3)
= -500(.9074)(.04979)
= -22.6

I know the answer for this question should not be negative as it deals with population regrowth. Where did I go wrong? Thanks.
$\displaystyle (1+e^{-t})'$ does not equal $\displaystyle (e^{-t})ln e$

4. Originally Posted by Jools
Got a question here....

Find the derivative of :

P(t) = 500 / (1 + e^-t)

I started by re-writing it in the form :

P(t) = 500 (1 + e^-t)^-1

Using the chain rule I came up with :

P'(t) = 500(-1)(1 + e^-t)^-2 (1 + e^-t)'

500(-1)(1+e^-t)^-2 (-(e^-t)) = 500 (1+e^-t)^-2 (e^-t)

The question goes on to ask for the derivative at P'(3)

P'(3) = -500(1 + e^-3)^-2(e^-3)
= -500(.9074)(.04979)
= -22.6
Since the derivative is 500(1 + e^-3)^-2(e^-3), your value should not come up negative now
I know the answer for this question should not be negative as it deals with population regrowth. Where did I go wrong? Thanks.
The derivative of $\displaystyle \frac{500}{1 + e^{-t}}$ is $\displaystyle \frac {500{e^{-t}}}{(1+e^{-t})^2}$

5. Would I also be correct in writing it like this:

-500(1 + e^-t)^-2 (-t e^-t-1)

Then solving for P '(3)?

It doesn't give the same answer when I solve for P '(3) as I get with your method, so I am assuming no...